We define a sequence of polynomials $\left( A _ { n } \right) _ { n \in \mathbb { N } } = \left( A _ { n } ( X ) \right) _ { n \in \mathbb { N } }$ satisfying the following conditions $$A _ { 0 } = 1 , A _ { n + 1 } ^ { \prime } = A _ { n } \text { and } \int _ { 0 } ^ { 1 } A _ { n + 1 } ( t ) d t = 0 \text { for every } n \in \mathbb { N }$$ The polynomials $B _ { n } = n ! A _ { n }$ are called Bernoulli polynomials.
a) Show that the sequence $\left( A _ { n } \right) _ { n \in \mathbb { N } }$ is uniquely determined by the conditions above; specify the degree of $A _ { n }$; calculate $A _ { 1 } , A _ { 2 }$ and $A _ { 3 }$.
b) Show that $A _ { n } ( t ) = ( - 1 ) ^ { n } A _ { n } ( 1 - t )$ for every $n \in \mathbb { N }$ and every $t \in \mathbb { R }$.
c) For every integer $n \geqslant 2$, show that $A _ { n } ( 0 ) = A _ { n } ( 1 )$ and that $A _ { 2 n - 1 } ( 0 ) = 0$.
d) We provisionally set $c _ { n } = A _ { n } ( 0 )$ for every natural integer $n$. Show that for every $n \in \mathbb { N }$, $$A _ { n } ( X ) = c _ { 0 } \frac { X ^ { n } } { n ! } + \cdots + c _ { n - 2 } \frac { X ^ { 2 } } { 2 ! } + c _ { n - 1 } X + c _ { n }$$ then that if $n \geqslant 1$, $$\frac { c _ { 0 } } { ( n + 1 ) ! } + \cdots + \frac { c _ { n - 2 } } { 3 ! } + \frac { c _ { n - 1 } } { 2 ! } + c _ { n } = 0$$
e) Deduce that for every $n \in \mathbb { N }$, we actually have $c _ { n } = a _ { n }$.